# Polarisation destroying interference pattern for double slit

• andrewkirk
In summary, the author is trying to explain why, despite using orthogonal polarisers, an interference pattern is still observed when light is passed through a double slit. He concludes that he must have made a mistake in his analysis.
andrewkirk
Homework Helper
Gold Member
I have been trying to mathematically explain the empirical result that putting orthogonal polarisers (quarter-wave plates) behind the two slits of a double-slit setup will erase the interference pattern.

The trouble is, my analysis predicts an interference pattern. I must have made a silly mistake, but I can't spot it. I was hoping somebody could help me find and correct it.

Here goes:

Assume monochromatic, coherent, linearly polarised light with polarisation vector ##\vec{x}## is directed towards a double slit. By definition ##\vec{x}## is the direction of the vector of maximal electric (magnetic) field intensity associated with each photon. Let ##\vec{y}## be the direction of maximal magnetic field intensity of photon. Both ##\vec{x}## and ##\vec{y}## are perpendicular to the direction of propagation. A quarter-wave plate is placed behind each slit, perpendicular to the direction of propagation. The fast axes of the plates behind slits 1 and 2 have directions ##\vec{u}\equiv\frac{1}{\sqrt{2}}(\vec{x}+\vec{y})## and ##\vec{v}\equiv\frac{1}{\sqrt{2}}(\vec{x}-\vec{y})## respectively, so that ##\vec{x}=\frac{1}{\sqrt{2}}(\vec{u}+\vec{v})## and ##\vec{y}=\frac{1}{\sqrt{2}}(\vec{u}-\vec{v})##. The effect of the plates is to impose a half-period phase delay of on the component of the polarisation vector aligned with the slow axis. This circularly polarises photons passing through the two slits, in opposite directions.

Because the light is coherent, every photon that has been emitted but not yet entered the plates has a maximum electric field in its polarisation direction at the same time. Let the electric and magnetic field intensity of all photons, in the period between emission and striking a plate, be given by ##A\sin kt##. That is, the electric and magnetic field vectors are ##\vec{x}A\sin kt## and ##\vec{y}A\sin kt## respectively.

Then, for a photon that has passed through the first slit, the electric field at time ##t## has components ##A\sin kt## and ##A\sin(kt-\frac{\pi}{2})=-A\cos kt## in directions ##\vec{u}## and ##\vec{v}## respectively. Hence the electric field after passing the plate has constant strength ##A## and the direction steadily rotates as ##\vec{u}\sin kt-\vec{v}\cos kt##. The normalised polarisation vector is ##\vec{p}_1(t)\equiv\frac{1}{\sqrt{2}}(\vec{u}\sin kt-\vec{v}\cos kt)##.

For a photon that has passed through the second slit,the electric field at time ##t## has components ##-A\cos kt## and ##A\sin kt## in directions ##\vec{u}## and ##\vec{v}## respectively. Again the electric field after passing the plate has constant strength ##A## and rotates with direction being ##-\vec{u}\cos kt+\vec{v}\sin kt##. The normalised polarisation vector is ##\vec{p}_2(t)\equiv\frac{1}{\sqrt{2}}(-\vec{u}\cos kt+\vec{v}\sin kt)##.

If the photon reaches the far field screen at time ##t##, it has electrical field of strength ##A## pointing in direction ##\vec{p}_1(t)## or ##\vec{p}_2(t)## according to which slit it passed through.

The time taken for a photon to reach a position ##\vec{r}## on the screen will depend on the distance from the slit it has passed through to the point ##\vec{r}##. For every point ##\vec{r}## there will be a difference ##d(\vec{r})## by which the distance traveled via the first slit exceeds the distance traveled via the second slit, to that point. The difference may be negative. Hence there will be a difference ##\theta(\vec{r})\equiv\frac{k\,d(\vec{r})}{c}## in the phases of (sine and cosine functions of) photons reaching that point.

Hence, at time ##t##, the electrical field at point ##\vec{r}## on the screen will be (ignoring the diminution of intensity by distance):

\begin{align*}
\frac{A}{\sqrt{2}}(\vec{u}\sin kt-\vec{v}\cos kt) &+ \frac{A}{\sqrt{2}}\big(-\vec{u}\cos (kt+\theta)+\vec{v}\sin (kt+\theta)\big)\\
&=\frac{A}{\sqrt{2}}\big(
\vec{u}\sin kt-\vec{v}\cos kt
-\vec{u}(\cos kt\cos\theta-\sin kt\sin\theta)
+\vec{v}(\sin kt\cos\theta+\cos kt\sin\theta)
\big)\\
&=\frac{A}{\sqrt{2}}\big(
\vec{u}
(\sin kt
-\cos kt\cos\theta+\sin kt\sin\theta)
+\vec{v}(-\cos kt
+\sin kt\cos\theta+\cos kt\sin\theta)
\big)\\
\end{align*}

Since ##\vec{u}## and ##\vec{v}## are orthogonal, the magnitude of this is:

\begin{align*}
\frac{A^2}{2}\big(
(&\sin kt
-\cos kt\cos\theta+\sin kt\sin\theta)^2
+(-\cos kt
+\sin kt\cos\theta+\cos kt\sin\theta)^2\big)\\
&\propto
\sin^2 kt
+\cos^2 kt\cos^2\theta+\sin^2 kt\sin^2\theta
-2\sin kt\cos kt\cos\theta
-2\cos kt\sin kt \cos\theta\sin\theta
+2\sin^2 kt\sin\theta\\
&\ \ \ \ \
+\cos^2 kt+\sin^2 kt\cos^2\theta+\cos^2 kt\sin^2\theta
-2\cos kt\sin kt\cos\theta
+2\sin kt\cos kt\cos\theta\sin\theta
-2\cos^2 kt\sin\theta\\
&=2
-4\cos kt\sin kt\cos\theta
+2(\sin^2 kt-\cos^2 kt)\sin\theta\\
&=2
-2\sin 2kt\cos\theta
-2\cos 2kt\sin\theta\\
&\propto 1-\sin(2kt+\theta)
\end{align*}

This will vary sinusoidally by ##\vec{r}## and hence will give an interference pattern.

What have I done wrong? There should be no interference pattern because the photons have been marked with which-way information by the differently oriented plates behind the two slits. thank you in advance for any help.

qopdaniela
Dr. Courtney said:
That looks like the answer to a different, and simpler, question. It is about whether two waves with orthogonal linear polarisations interfere. It is very easy to see that they do not.

My question is about where the waves have circular polarisations of opposite senses.

Having received no solutions after three days, I posted this on Stack Exchange and got an answer from Kevin Zhou that solved it.

What I was doing wrong was forgetting to do the last step, which is to integrate the magnitude (squared field intensity) over time to get time-average square intensity - which is all we can observe since the period is so short. If the result is independent of ##\theta## there will be no interference pattern. If it is a periodic function of ##\theta## there will. In the above case, the time-average square intensity is

$$\frac{1}{\frac{\pi}{k}}\int_0^{\frac{\pi}{k}}\big(1-\sin (2kt+\theta)\big)dt=1$$

So it is independent of ##\theta## as required. Hence there is no interference pattern.

whitsona2

## 1. What is a polarisation destroying interference pattern for a double slit?

A polarisation destroying interference pattern for a double slit is a phenomenon that occurs when light waves with different polarisations interfere with each other after passing through two parallel slits. This interference results in a pattern of light and dark fringes on a screen placed behind the slits.

## 2. How does polarisation destroying interference occur?

Polarisation destroying interference occurs when light waves with different polarisations pass through two parallel slits and interfere with each other. The polarisation of the light waves is altered as they pass through the slits, resulting in changes to the direction of the electric and magnetic fields. These changes cause the light waves to interfere with each other and produce a pattern on the screen.

## 3. What factors affect the polarisation destroying interference pattern?

The polarisation destroying interference pattern can be affected by several factors, including the distance between the slits, the wavelength of the light, and the angle at which the light waves pass through the slits. Additionally, the type of material used for the slits and the polarisation of the incident light can also affect the pattern.

## 4. Can the polarisation destroying interference pattern be manipulated?

Yes, the polarisation destroying interference pattern can be manipulated by changing the angle of the incident light or by using polarising filters. By adjusting these factors, the resulting interference pattern can be altered or even eliminated.

## 5. What are the practical applications of polarisation destroying interference for double slits?

Polarisation destroying interference for double slits has various practical applications, including polarisation-based imaging techniques in microscopy, optical filtering and polarimetry, and polarisation-sensitive detectors in certain scientific instruments. This phenomenon also plays a crucial role in understanding the properties of light and can be used to study the behaviour of polarised light waves.

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